Consider the parabola $y^2=4x$. Let $V$ be the vertex, $F$ be the focus and $P$ be a point on the parabola such that $T$ is the foot of perpendicular of point $P$ on the directrix. Find a point $P$ such that the distance between $V$ and the orthocenter of $\Delta PFT$ is minimized. Find this value.
Thoughts:
1) When $P$ approaches $V$, $\angle FPT$ goes on increasing approaching $\pi$ radians(triangle being obtuse). The orthocenter in this case goes off to infinity.
2) When $P$ approaches infinity on either branch of the parabola, the orthocenter of this acute-angled triangle also approaches infinity(not sure about this one)
So I naively concluded that the triangle should be right-angled(at $P$) for this optimization(turns out that this is actually correct, verified using geogebra).
However, this has little to no rigor. I'm looking for comments on my thoughts and better approaches to this problem.
Here's the graph with a slider
You are right that the minimum distance from vertex to orthocenter occurs when $\triangle PFT$ is right angled triangle (isosceles, of course) with $\angle FPT = 90^0$.
I will present two solutions -
a) First solution:
Using reflective property of parabola, tangent at $P$ bisects $\angle FPT$ and as $PT = PF$, the tangent at $P$ is also perpendicular bisector of $FT$. Next, the vertical line at $F$ (line $x = 1$) is perpendicular to $PT$. So the orthocenter of the $\triangle FPT$ must be the intersection of the tangent at $P$ and line $x = 1$, say point $X$. As a tangent never intersects parabola at two points, point $X$ must be outside the parabola or on the parabola, when both $P$ and $X$ are at point $P'$, which is intersection of $x = 1$ with the parabola itself.
As the distance from $V$ to $P'$ is lower than distance from $V$ to any other point on line $x = 1$ outside the parabola, we have point $P' (1, \pm 2)$ as our answer.
b) Second solution:
Parabola is $y^2 = 4x$. Say, point P is $(p^2, 2p), \ p \in \mathbb{R}$
Vertex $V$ is $(0, 0)$, Focus $F$ is $(1, 0)$ and directrix is $x = -1$.
Coordinates of $T$ is $(-1, 2p)$.
In $\triangle PFT$, it is easy to see that orthocenter lies on $x = 1$.
As $PT = PF$, $\triangle PFT$ is isosceles and if $M$ is the midpoint of $FT$, $PM \perp FT$.
Given coordinates of $F$ and $T$, coordinates of $M$ is $(0, p)$.
Equation of line going through $PM$,
$y - p = \frac{2p-p}{p^2-0} (x-0) \implies y - p = \frac{x}{p}$ if $p \ne 0$
Intersection of line $x = 1$ and $PM$ is the orthocenter, which is $X \left(1, p + \frac{1}{p}\right)$.
$(VX)^2 = \left(p + \frac{1}{p}\right)^2 + 1 = p^2 + \frac{1}{p^2} + 3$
By AM-GM, min value of $p^2 + \frac{1}{p^2}$ is $2$, when $p^2 = 1$
So the points on parabola that minimize distance $VX$ are $(1, \pm2)$ when both $P$ and $X$ coincide at point $P'$.